test2solf05

test2solf05 - ‘l " Physics 113, Fall, 2005 Exam...

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Unformatted text preview: ‘l " Physics 113, Fall, 2005 Exam 2 Tuesday Nov. 15, 2005 NAME Workshop Leader 6 SN Directions 0 Answer all questions. The test has 100 total points. 0 You must show all work in order to receive full credit. 0 No books are allowed. One fact sheet is provided (See last page.) You may use calculators. 0 Put your name on EACH PAGE. 0 If you cannot answer the question in the space allotted, ask for a separate piece of paper. 0 SHOW ALL WORK. NAME 1) (25 points total) A child (m = 25 kg) slides down a play slide with 6 = 35° and with coefficient of kinetic friction of me = 0.2. (See figure). / mzzskg. - *5 1A) 5 Points V Draw a freebody diagram for the child. E (“an”) FM (named) " 3) " NAME 1A) 20 Points If the child starts from rest at the top of the slide, d = 4m above the bottom, how fast will she be traveling when she reaches the bottom? F’Hé’fiin doco Lac-ML (Wm work) 6‘0 6t 00"“ chQeAVGV’WC game— \K ’sto Pmkw. UK (yak: CWU5\1 rzkahmspr: Wflilt: AKEKT 2<l(.€.€';4* > + (Rated ’9545M> Z flimqyso ire.\\$ LS HA8? FU—mj cams) :0 ' .- Fm; — 3 3 5643‘) (Q cms\rav3f {W‘Q (DC-W“) 9Q M h? fl FN fimac . S\\'0Lo.> 5M d\rto’nb\~5 \Un L 3 74mme (A smka d’v Re arule 0096 2) 25 Points The ballistic pendulum is a device _used to measure the speed of a bullet. The bullet of mass m1 is fired into a large block of wood of mass m2 which is suspended like a pendulum. The bullet is embedded in the block of wood. As a result of the collision, ‘ the pendulum swings up to a maximum height h. Find an expression for the initial velocity of the bullet v in terms of the height h. block me Massimojl’lq piwfi‘ cam/{oi {Xer’i q (pro. ovx Mblook dorivxj mCO\\\s\on, SO 11" \lw dova bailéve \“MsJ ‘10» can (AC - macho/j 0mm? Mama'be MA by» L~\\\ Swims M12. Sirlhty holde 3111 .5 is cans-C(U‘CA. ( Qn6L\0r (“0wa (0/56 956* 4%. saw (650*) . 7P? 3 i: . miv = w; (an m Axolireblim - l0 3 W‘fi’ml V SW9- r f ) 4F ' 0am C5\\\$)1026}LQ 50 d (“N :(mcl'mvlvs: 3D CO d medium medemm iscmsemid, fir EL ii: kgmemm/j __ 5-_ NAME 3) '(25 points total) A uniform m = 100 kg disk with radius r = 0.6m is placed flat on a sooth ice—skating rink (assumed to be frictionless). Two skaters wind ropes around the cylinders in the same sense (counter-clockwise). The skaters then pull on their ropes as they skate away, exerting forces of F1 :40 N and F2 = 60 N, r we +52 espectively, on the disk for 5 s. 3A) 12 Points Find the position of the center of mass of the disk as a function of time during the 5 seconds. Be sure to clearly indicate your coordinate system. The com. otm disk-Wklmim cattle COW/r Olflt‘ dusk wouth ms Pman _ ( what we‘d flformw olile . 0w) change its com?!) LU€ (pant in find w‘mm‘t‘ne (ML uu{\\b€. a0 4 finchim 6% . TN. (JV-\Sk has a fig chbcdomdwk £01m 0A itmt ' . - . 7 m . gov L1 oVU mm am mane cam w: 1M +x amino/t a: 7:0)" ~ 0.7. W51 lutuhéwt . ’l Comm accwahm [We xxm Watt ' \’s COMM \vk Sew/win- - é - NAME 3B) 13 Points > Find the angular acceleration and angular “velocity of the disk as a. function of time during the 5 seconds. ‘00?ij down Ohm disk 14’me \vp1 E L40“) / VJ A F7. tw: HRJVFL'R Goa/h dwkufiej Cam-BC TdYELV) f¢*an=(‘4Mr&M " 1U 0L: 1(F\W~*F7'¥) : Zaéfi F2. 200 W MK (lwkj)[0‘ém) x r: 333 «4/51 3 a comymf LA)... 5° ‘ (Ravi : 333% \A (Ms/5a {f JciS I o (39th \\-r\ Scwndp —- q— — NAME 4) ( 25 points) I A uniform trap door of mass m = 5kg is l = 0.75m wide, and hinged on one side, as shown. It is held open at an angle 6 = 55° by a massless cable that makes a right angle with the trap door. 4A) 9 Points Find the Tension (T) in the cable. //- llle "lckqw (4le N l'lwlor {Ci/C.sz T “Q 35: TX.- rfifl/COSQ) T; w C015, : S‘(‘i.8)('.s%): MW 8 NAME 48) 9 Points Find the force on the trap door at the hinge. q NAME 40) 7 Points The massless cable snaps, allowing the trapdoor to swing shut. Find the velocity of the edge of the trapdoor farthest from the hinge as it hits the floor. It might be useful to know that the moment of inertia. of a board of length l and mass M pivoted about its center is I = fiMl2 and a board pivoted about one end is I = %M 12. L) Sc. enact}, comerth 95ch I; no\’ cooéhhn‘l'. 15;: mfihlwterc h Mu pos'mm of m cé/Ln. . EL 1 . mahcm: “M916”. 7.. < ' l o -’; , ' 1 no KYVW‘L JrCrwx b60943le [Fri Llw (’nro‘m’imj/no’f translmfihf) ) ...
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test2solf05 - ‘l &amp;quot; Physics 113, Fall, 2005 Exam...

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